The equation f (x) = 4x^2 - 12x + 13 is written as the equivalent function f (x) (2x-3)^2 + k. What is the vale of k?
I got an answer from someone else, and I'm not quote sure if I understood their answer clearly, but I got 9? Is it correct? This is very important because I have to finish this benchmark, and then I have to finish another and I only have a few hours left to do it! I really need help!
square the binomial and compare the results to find k
(2x-3)^2 = 4x^2 - 12x + 9
(2x-3)^2 + k = 4x^2 - 12x + 13
check your calculations
4 x² - 12 x + 13 = ( 2 x - 3 )² + k
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Remark:
( m - n )² = m² - 2 ∙ m ∙ n + n²
( 2 x - 3 )² = ( 2 x )² - 2 ∙ 2 x ∙ 3 + 3² + k
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4 x² - 12 x + 13 = ( 2 x )² - 2 ∙ 2 x ∙ 3 + 3² + k
4 x² - 12 x + 13 = 4 x² - 12 x + 9 + k
Subtract 4 x² - 12 x + 9 to both sides
4 x² - 12 x + 13 - ( 4 x² - 12 x + 9 ) = ( 4 x² - 12 x + 9 ) + k - ( 4 x² - 12 x + 9 )
4 x² - 12 x + 13 - 4 x² + 12 x - 9 = k
4 x² - 4 x² - 12 x + 12 x + 13 - 9 = k
4 = k
k = 4
4 x² - 12 x + 13 = ( 2 x - 3 )² + 4
Thank you!
F(x) = 4x^2 - 12x + 13.
h = -B/2A = 12/8 = 3/2.
K = 4(3/2)^2 - 12(3/2) + 13 = 9 - 18 + 13 = 4.
Vertex Form: F(x) = a(x-h)^2 + k. = 4(x - 3/2)^2 + 4.
To find the value of k in the equation f(x) = (2x-3)^2 + k, you can compare this equation with the given equation f(x) = 4x^2 - 12x + 13.
In the given equation, we can see that the constant term is 13. This is the term that doesn't have an x variable. In the equation f(x) = (2x-3)^2 + k, the constant term is represented by k.
So, in order to find k, we need to set the constant term from the given equation equal to k.
Setting the constant term equal to 13, we have:
k = 13
Therefore, the value of k is 13, not 9 as you mentioned.